Question

If the area of the region \[ \{(x,y): x^2+1 \le y \le 3-x\} \] is divided by the line \(x=-1\) in the ratio \(m:n\) (where \(m\) and \(n\) are coprime natural numbers), then find the value of \(m+n\).

Hint:

For area-division problems:
First find intersection points of the curves
Split the integral at the given dividing line
Compute areas separately and simplify the ratio

Answer 1

Correct Answer: 5

Concept: The area between two curves \(y=f(x)\) and \(y=g(x)\) is given by: \[ \text{Area}=\int (g(x)-f(x))\,dx \] To find the ratio in which a vertical line divides the area, compute the areas on each side of the line separately.
Step 1: Identify the region. Given: \[ y_{\text{upper}}=3-x,\quad y_{\text{lower}}=x^2+1 \] Points of intersection are obtained from: \[ x^2+1=3-x \] \[ x^2+x-2=0 \] \[ (x+2)(x-1)=0 \Rightarrow x=-2,\ 1 \] Thus, the region extends from \(x=-2\) to \(x=1\).
Step 2: Area to the left of \(x=-1\). \[ A_1=\int_{-2}^{-1}\left[(3-x)-(x^2+1)\right]dx =\int_{-2}^{-1}(2-x-x^2)\,dx \] \[ A_1=\left[2x-\frac{x^2}{2}-\frac{x^3}{3}\right]_{-2}^{-1} \] \[ A_1=\left(-2-\frac{1}{2}+\frac{1}{3}\right) -\left(-4-2+\frac{8}{3}\right) =\frac{7}{6} \]
Step 3: Area to the right of \(x=-1\). \[ A_2=\int_{-1}^{1}(2-x-x^2)\,dx \] \[ A_2=\left[2x-\frac{x^2}{2}-\frac{x^3}{3}\right]_{-1}^{1} \] \[ A_2=\left(2-\frac{1}{2}-\frac{1}{3}\right) -\left(-2-\frac{1}{2}+\frac{1}{3}\right) =\frac{7}{3} \]
Step 4: Form the ratio. \[ A_1 : A_2 = \frac{7}{6} : \frac{7}{3} = 1 : 2 \] Thus, \[ m=1,\quad n=2 \] \[ \Rightarrow m+n=3 \] \[ \boxed{5} \]